Problem: You are given an integer N.

Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.

* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.

Constraints

* 2 \leq N \leq 500

Input

Input is given from Standard Input in the following format:


N


Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Example

Input

2


Output

4 7
23 10

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define int long long
using namespace std;
const int N = 1e3 + 5;
int read() {
	int x = 0, f = 1; char ch;
	while(! isdigit(ch = getchar())) (ch == '-') && (f = -f);
	for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 3) + (x << 1) + (ch ^ 48));
	return x * f;
}
template <class T> T Max(T a, T b) { return a > b ? a : b; }
template <class T> T Min(T a, T b) { return a < b ? a : b; }
int n, tot, vis[1000005], pri[1000005], a[N][N];
void init(int n) {
	for(int i = 2; i <= n; ++ i) {
		if(vis[i] == 0) pri[++ tot] = i;
		for(int j = 1; j <= tot && i * pri[j] <= n; ++ j) {
			vis[i * pri[j]] = 1;
			if(i % pri[j] == 0) break;
		}
	}
}
int gcd(int x, int y) {return y == 0 ? x : gcd(y, x % y);}
int lcm(int x, int y) {
	if(x == 0 || y == 0) return x + y;
	return x / gcd(x, y) * y;
}
signed main() {
	n = read(); init(1e4);
	if(n == 2) return puts("4 7\n23 10"), 0;
	for(int i = 1; i <= n; ++ i) {
		for(int j = ((i + 1) & 1) + 1; j <= n; j += 2) {
			a[i][j] = pri[(i + j) / 2] * pri[n + (i - j) / 2 + (n + 1) / 2];
		}
	}
	for(int i = 1; i <= n; ++ i) {
		for(int j = (i & 1) + 1; j <= n; j += 2) {
			a[i][j] = lcm(lcm(a[i - 1][j], a[i + 1][j]), lcm(a[i][j - 1], a[i][j + 1])) + 1;
		}
	}
	for(int i = 1; i <= n; ++ i, puts("")) for(int j = 1; j <= n; ++ j) printf("%lld ", a[i][j]);
	return 0;
}